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Replacing an operator of an angular momentum for a constant.

  1. Jun 5, 2013 #1
    While dealing with a circling particle in an spherical symetric potential our professor said that we can replace an operator of ##z## component of angular momentum ##\hat{L}_z## with the expectation value - he denoted it just ##L_z## - of the angular momentum if ##L_z## is constant. Why is that?

    So we first had this equation:

    \begin{align}
    \underbrace{\psi (r,\varphi,\vartheta)}_{\rlap{\text{w. f. in spherical coordinates}}} &= \exp\left[\hat{L}_z \frac{i}{\hbar}\, \varphi\right] \underbrace{\psi (r,0,\vartheta)}_{\rlap{\text{w. f. in spherical. coordinates at $\varphi=0$}}}
    \end{align}

    and we got this one (notice that there is no operator over an ##L_z##):

    \begin{align}
    \psi (r,\varphi,\vartheta) &= \exp\left[L_z \frac{i}{\hbar}\, \varphi\right] \psi (r,0,\vartheta)
    \end{align}

    Anyway here is the spherical coordinate system we ve been using all the time (the blue spherical aure is supposed to be a spherical potential...):

    G8aUE.png
     
    Last edited: Jun 5, 2013
  2. jcsd
  3. Jun 5, 2013 #2

    Jano L.

    User Avatar
    Gold Member

    I think there should be imaginary unit inside the[ ] brackets. The first equation states that the function ##\psi## is analytic in ##\varphi##. The second equation states more than that: it says that the function behaves as ##e^{im\varphi}##, i.e. oscillates. Such functions are eigenfunctions of the angular momentum operator, if the number ##m = l_z/\hbar## is whole number.
     
  4. Jun 5, 2013 #3
    You are soo right! I misstyped ##i## for ##1##. I corrected that mistake. I don't think i understand your explaination though.
     
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